A solid wooden toy is in the form of a cone mounted on a hemisphere. The radii of the hemisphere and the base of the cone are 4.2 cm each and the total height of the toy is 10.2 cm. Calculate :

(i) the volume of wood used in the toy

(ii) the total surface area of the toy, correct to two places of decimal.

Given,

The solid wooden toy is in the shape of a right circular cone mounted on a hemisphere.

Radius of hemisphere, r = 4.2 cm

Total height, h = 10.2 cm

Height of conical part, H = 10.2 - 4.2 = 6 cm

(i) Volume of wood used in toy = Volume of cone + Volume of hemisphere

$= \dfrac{1}{3} π\text{r}^2\text{H} + \dfrac{2}{3} π\text{r}^3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 4.2^2 \times 6 + \dfrac{2}{3} \times \dfrac{22}{7} \times 4.2^3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 17.64 \times 6 + \dfrac{2}{3} \times \dfrac{22}{7} \times 74.088 \\[1em] = \dfrac{2328.48}{21} + \dfrac{3259.872}{21} \\[1em] = \dfrac{5588.352}{21} \\[1em] = 266.11 \text{cm}^3$

Hence, the volume of wood used in the toy is 266.11 cm³.

(ii) By formula,

l² = r² + h²

⇒ l² = 4.2² + 6²

⇒ l² = 17.64 + 36

⇒ l² = 53.64

⇒ l = $\sqrt{53.64}$ = 7.32 cm

Total surface area of toy = Curved surface area of cone + curved surface area of hemisphere

$= π\text{rl} + 2π\text{r}^2 \\[1em] = \dfrac{22}{7} \times 4.2 \times 7.32 + 2 \times \dfrac{22}{7} \times 4.2^2 \\[1em] = \dfrac{676.368}{7} + 2 \times \dfrac{22}{7} \times 17.64 \\[1em] = \dfrac{676.368}{7} + \dfrac{776.16}{7} \\[1em] = \dfrac{676.368 + 776.16}{7} \\[1em] = \dfrac{1452.528}{7} \\[1em] = 207.56 \text{cm}^2$

Hence, the total surface area of the toy is 207.56 cm².